Internal problem ID [10121]
Internal file name [OUTPUT/9068_Monday_June_06_2022_06_21_13_AM_24431925/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1800 (book 6.209).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{3} y^{\prime \prime }=a} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} y^{3} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = a \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {a}{y^{3} p} \end {align*}
Where \(f(y)=\frac {a}{y^{3}}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {a}{y^{3}} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {a}{y^{3}} \,d y} \\ \frac {p^{2}}{2}&=-\frac {a}{2 y^{2}}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}+\frac {a}{2 y^{2}}-c_{1} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {{y^{\prime }}^{2}}{2}+\frac {a}{2 y^{2}}-c_{1} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {y}{\sqrt {2 c_{1} y^{2}-a}}d y &= \int d x \\ \frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {y}{\sqrt {2 c_{1} y^{2}-a}}d y &= \int d x \\ -\frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}} &= x +c_{2} \\ \tag{2} -\frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}} = x +c_{2} \] Verified OK.
\[ -\frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{3} \left (\frac {d}{d x}y^{\prime }\right )=a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),\frac {d}{d x}y^{\prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & y^{3} u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=a \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {a}{y^{3} u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {a}{y^{3}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int \frac {a}{y^{3}}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=-\frac {a}{2 y^{2}}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {2 c_{1} y^{2}-a}}{y}, u \left (y \right )=-\frac {\sqrt {2 c_{1} y^{2}-a}}{y}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y y^{\prime }}{\sqrt {2 c_{1} y^{2}-a}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y y^{\prime }}{\sqrt {2 c_{1} y^{2}-a}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}+a \right )}}{2 c_{1}}, y=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}+a \right )}}{2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {2 c_{1} y^{2}-a}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y y^{\prime }}{\sqrt {2 c_{1} y^{2}-a}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y y^{\prime }}{\sqrt {2 c_{1} y^{2}-a}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 c_{1} y^{2}-a}}{2 c_{1}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}+a \right )}}{2 c_{1}}, y=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}+a \right )}}{2 c_{1}}\right \} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-a/_a^3 = 0, _b(_a), HINT = [[_a, -_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, -_b]
✓ Solution by Maple
Time used: 15.657 (sec). Leaf size: 46
dsolve(y(x)^3*diff(diff(y(x),x),x)-a=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\sqrt {\left (\left (c_{2} +x \right )^{2} c_{1}^{2}+a \right ) c_{1}}}{c_{1}} \\ y \left (x \right ) &= -\frac {\sqrt {\left (\left (c_{2} +x \right )^{2} c_{1}^{2}+a \right ) c_{1}}}{c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 4.192 (sec). Leaf size: 63
DSolve[-a + y[x]^3*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {a+c_1{}^2 (x+c_2){}^2}}{\sqrt {c_1}} \\ y(x)\to \frac {\sqrt {a+c_1{}^2 (x+c_2){}^2}}{\sqrt {c_1}} \\ y(x)\to \text {Indeterminate} \\ \end{align*}